Workshop Notes

最後一次整理 workshop notes 是從 UG year 3 到 PG year 1 的暑假。最近懂得用 mtpro2 這字體，便從新再換一下格式：

http://ihome.ust.hk/~cclee/document/WorkshopinAnalysis.pdf

若我能如期在暑假前畢業，可能會再多開一次 workshop 去說說 complex analysis --- 集合 real, functional 及 harmonic analysis 的一門優美學科。

weak $C^1$ norm dominates $C^0$ norm on $C^1[0,1]$ (to be used in tutorial).

The weak $C^k$ norm for $f:[a,b] \to \R$ is defined by \[
\|f\|_{W^{k,1}} =\sum_{i=0}^k \int_a^b |f^{(i)}(t)|\,dt.
\] This is an exercise (a problem stated in some forum) to show that the weak $C^1$ norm (hence weak $C^k$ norm, this concept comes from Sobolev spaces) dominates the $C^0$ norm (i.e., sup-norm):

Problem. If $f\in C^1[0,1]$, show that \[
\sup_{x\ge\in [0,1]}|f(x)|\leq \int_0^1|f(t)|\,dt +\int_0^1 |f'(t)|\,dt.
\] 

Solution. Denote $I=[0,1]$. Let $x_0\in I$ be such that $|f(x_0)|=\|f\|_{[0,1]}$. Then for any $\alpha,\beta\in I$ we have \[
\int_{x_0}^\beta |f'(t)|\,dt \ge |f(x_0)-f(\beta)|
\] and \[
\int_{\alpha}^{x_0} |f'(t)|\,dt \ge |f(x_0)-f(\alpha)|.
\] On summing up, by triangle inequality we have \[
\int_0^1|f'(t)|\,dt \ge \int_\alpha^\beta |f'(t)|\,dt \ge |2f(x_0)-f(\alpha)-f(\beta)|.
\] On the other hand, by continuity of $f$ and the differentiability of $g(x):=\int_0^x|f(t)|\,dt$ on $(0,1)$ we have \[
\int_0^1|f(t)|\,dt =g(1)-g(0)=g'(c)=|f(c)|,
\] for some $c\in (0,1)$. Therefore by triangle inequality again we have \begin{align*}
\|f\|_{W^1[0,1]}&:= \int_0^1|f(t)|\,dt+\int_{0}^{1} |f'(t)|\,dt\\
&\ge  |2f(x_0)-f(\alpha)-f(\beta)| + |f(c)|\\
&\ge |2f(x_0)-f(\alpha)-f(\beta)+f(c)|.
\end{align*} Now we set $\alpha = x_0$ and $\beta=c$ to get $\|f\|_{W^1[0,1]}\ge |f(x_0)|=\|f\|_{[0,1]}$, as desired.$\qed$

教授安裝 mathtime pro 2 字型

tex stackexchange 那有教授超簡單的使用方法：
http://tex.stackexchange.com/questions/110894/how-do-i-install-mtpro2

\pdfmapfile{=mtpro2.map}

在 office 也能夠輕鬆使用，不用安裝！(以前為安裝這字體弄得死去活來，耗費不少青春！)

某文件

在接着的學期我將要教 Math2033 的 tutorial (即 201)，instructor 由教了好幾年的何漢明轉為 Dr kin li。Math 2031 (即 202) 暫定不會再出現。

kin li 的 tutorial 對非常懶的 TA 來說很好，kin li 不會要求你在 tutorial 教甚麼，只須坐在課室聽學生做每周安排的 presentation 便何。但在我以至我前兩屆師兄的年代 (kin li 的) Math202/301 tutorial 不會單純聽 presentation，會用盡剩餘的時期多講講要知的 concept 和例子。為了延續這份精神，我不曾浪費掉 Math301 tutorial 多出的時間，這次的 Math2033 當然也不例外。因此我想起這份 pdf：

http://ihome.ust.hk/~cclee/document/Questions.pdf

中學時期的我便會使用 tex。那時很迷 tex，typeset 的數式很美，所以非常勇躍去解有趣的數學題，把它的解打出來，然後收錄。這習慣只維持到大學二年級。百多頁的文件處理起來不容易。就算現在懂得把 tex file 局部處理再合併 (tex 裏面做) 也懶得再整理這文件了。因我解的題愈來愈多，把我大學三年級到 mphil year 2 裏我認為有趣的題都收錄的話這份文件恐怕頁數將超過 200 --- 不易整理。

但解過的題都總想記錄一下，現在的途徑多把題分散在自己的 tutorial notes 或在這 blog 裏。在以上連結裏 elementary analysis 的部分將會散落到 2033 的 tutorial notes 吧～ :)

呀，另外在某一年為 Math201 的學生炮製了某 list 問題：

http://ihome.ust.hk/~cclee/document/math201.pdf

Math3033 (2013-2014) Final Exam

This year the mean of the final exam is 35.51/100 and the SD is 12.71. To wit, answering the first two problems perfectly is already safe to get $\text{C+}/\text{B-}$. If one does perfectly well in everything before the final exam (which are all very standard), then one gets an $\text{A-}$. This again reflects that there are too many weak students in Math3033 and getting an $\text{A-}$ is very easy for hard working students.

The first two problems are routine calculation, with one problem on uniform convergence and one on Lebesgue Dominated Convergence Theorem. The next 3 problems are as follows:

Problem 3. Show that the function $f:[0,1]\setminus \Q\to \R$ given by \[
f(x)=\sum_{i=1}^\infty \frac{a_i}{16^i},\quad\text{where } x=0.a_1a_2a_3\dots
\] is measurable.

Summary to Problem 3. Very few people can get it correct, some partial marks (up to 2) are given for those who try to prove $f$ is continuous, increasing, etc. Indeed this function is not continuous, any attempt proving this to be continuous just deserve very little marks. Some of the students claim the function is increasing, this still requires some manipulations of inequality and cannot be seen as trivial.$\qed$

Problem 4. Let $f:[0,1]\to[0,\infty)$ be measurable such that \[
\int_{[0,1]} f^n\, dm =c\in \R,\quad \forall n\in \N.
\] (a) Prove that $f^{-1}[2013,\infty)$ has measure zero.
(b) Prove that $f=\chi_E$ a.e. for some measurable set $E$.

Summary to Problem 4. By observing the answer in part (b), some students can guess the answer in part (a) but did not attempt to prove this. Actually it is very easy to show $m(f^{-1}(\alpha,\infty))=0$ for every $\alpha > 1$, therefore $f\leq 1$ a.e. The part (b) is hard to many students, they need to explain why $m(f^{-1}[0,1))=0$, this should not be difficult if one tries to play around with the term $\int_{[0,1]}f^n\,dm = c$ and split the integral into two parts. Then \[
m(f^{-1}(\{1\}))=1,
\] and therefore on $f^{-1}(\{1\})$ we have $f = 1=\chi_{f^{-1}(\{1\})}$. It follows that $f=\chi_E$ a.e., where $E=f^{-1}(\{1\})$.$\qed$

Problem 5. Let $T\subseteq [0,1]$ be a set of $7$ elements and $S\subseteq [0,1]$ be measurable such that $m(S)\ge \frac{1}{3}$, prove that there are $a,b\in T$ and $c,d\in S$ such that \[
|a-c|=|b-d|.
\] 

Summary to Problem 5. Only 2 people can get a decent amount of marks from me. Both of them try to prove by contradiction by supposing that $|a-c|\neq |b-d|$ for every $a,b\in T$ and $c,d\in S$. One of them can proceed successfully while another one is on the half way. Students usually work on equivalent statement of $|a-c|=|b-d|$ but never think of the possibility of having the sufficient condition: \[
a-c=b-d
\] for some $a,b\in T$ and $c,d\in S$. In this way one can find that \[
a+d=b+c.
\] Namely, we may try to prove $(a+S)\cap( b+S)\neq \emptyset$ for some $a,b\in T$. Suppose not, the we have the contradiction that $2\ge 7/3$.$\qed$

Conclusion. It seems that Dr Li thinks the midterm is easy enough to ensure most people can pass the course, therefore it tries to set the exam in a moderate level. Unfortunately very few students can develop a good background (absorb the material) and technique (methodology) in the basic measure theory.

Fourier Series

Usual exercises in Fourier series ask students to prove identity of the form \[
f=\sum a_n\cos nx+\sum b_n\sin nx.
\] Most of them are extremely straightforward and what you need is sufficient manpower, patience and time. If I were the one who teach Fourier series, I would propose the following identities as examples:

Problem. By using the Fourier series expansion of $f(x)=\cos \alpha x$, $x\in (-\pi,\pi)$, prove that \[
\cot x=\frac{1}{x}+\sum_{n=1}^\infty \brac{\frac{1}{x-n\pi}+\frac{1}{x+n\pi}}=\frac{1}{x}+\sum_{n=1}^\infty \frac{2x}{x^2-n^2\pi^2}
\] and also \[
\csc x=\frac{1}{x}+\sum_{n=1}^\infty (-1)^n \brac{\frac{1}{x-n\pi} +\frac{1}{x+n\pi}}=\frac{1}{x}+\sum_{n=1}^\infty (-1)^n \frac{2x}{x^2-n^2\pi^2}.
\] Also evaluate $\dis \sum_{n=1}^\infty \frac{1}{n^2-\alpha^2}$.

The surprise should be: there is apparently no evidence that Fourier series is involved!

Integral Inequality

Problem. Let $k$ be a positive integer.  $f:[0,\infty)\to[0,\infty)$ is a continuous function such that $f(f(x))=x^k$, $\forall x\in[0,\infty)$. Show that \[\int_0^1f(x)^2\,dx\ge\frac{2k-1}{k^2+6k-3}.\]

This statement is provable by Young's inequality. By assuming $f$ be continuously differentiable (or any kind of continuous such that the integration by parts on $[0,1]$ work, say $f$ is absolutely continuous), then we have a stronger bound: \[
\int_0^1f(x)^2\,dx\ge\frac{2k-1}{k^2+4k-2}.
\] Which becomes the previous one by adding $2k-1$ to the denominator.

Solution. There is a few observations. First, $f$ is injective. Second, $f(f(x))=x^k$ implies $f(f(f(x)))=f(x)^k$, therefore  $f(x^k)=f(x)^k$. In particular, if we put $x=0$ and $x=1$, then we have $f(0)=f(0)^k$ and $f(1)=f(1)^k$. Therefore both $f(0),f(1)\in \{0,1\}$.  By injectivity and the fact $f\ge 0$, we have $f(0)=0$ and $f(1)=1$.

By Cauchy-Schwarz inequality we have $I:=\int_0^1f^2\,dx\ge \int_0^1 f\,dx$. We substitute $x=f(y)$ to get \[
I=\int_0^1 f(f(y))(f(y))'\,dy=\int_0^1 y^k(f(y))'\,dy=1-k\int_0^1f(y)y^{k-1}.
\] The last equality follows from integration by parts. Now we shrink RHS by Cauchy-Schwarz inequality to get $I\ge1- \frac{k}{\sqrt{2k-1}}\sqrt{I}$, which we rearrange to get \[
(\sqrt{I})^2 +\frac{k}{\sqrt{2k-1}} \sqrt{I}-1\ge 0.
\] By completing square we find that this implies \[
\sqrt{I}\ge \frac{-\frac{k}{\sqrt{2k-1}}+ \sqrt{\frac{k^2+8k-4}{2k-1}}}{2}=\frac{2\sqrt{2k-1}}{\sqrt{k^2+8k-4}+k}.
\] Now by Cauchy-Schwarz inequality  we also have \[
\sqrt{k^2+8k-4}+k\leq \sqrt{(k^2+8k-4+k^2)(1+1)}=2\sqrt{k^2+4k-2},
\] hence we have \[
I\ge \frac{2k-1}{k^2+4k-2}.\qed
\]

Code

在 inline mdoe 時 matrix 縮細，display mode 時 matrix 為原本大小的方法：

msc 好恐怖

近排食下午茶撞到 Dr. Li，咁就一齊傾下計。佢話我知原來 Jimmy fung 又話我俾人投訴喺 msc (math support center) 到爆左 d 單字出黎。唔知係咪 Dr. Fung 親耳聽到，定係有人投訴，我自己就已經愈黎愈小心，亦都甚少落 msc (落親去都係搵人食飯，唔會逗留多過十分鐘)，都仲要咁岩俾人捉到又咁岩俾人投訴。話冇人針對我，你信唔信？但我自己又心虛，難以否認。

Dr. Li 叫我盡量將所有精神放喺數學，唔好諗其他野，咁就冇咁易有野俾人捉住黎講，仲叫我盡量做到冇野可以俾人拎黎講。咁有 d 難度喎...。

一條毒 L 行淘大商埸

以前淘大商場 3 樓周圍都係賣電現模型漫畫多，而家有好多地方都變左做賣電子產品、飾品及補習，仲要有 759 零食部 ...。

以前呢到一上樓梯就係賣遊戲/遊戲機，玩具，遊戲卡等。果時好多細佬渣住部 gameboy 聚喺塊玻璃後面搵人玩 pokemon 對戰 ...。

同一層，上圖所示嘅鋪位對面就係下面呢個景像，以前呢到前、左、右都係賣遊戲/動畫相關嘅野，而家變哂飾品/家具...。